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प्रश्न
Show that the general solution of the differential equation `dy/dx + (y^2 + y +1)/(x^2 + x + 1) = 0` is given by (x + y + 1) = A (1 - x - y - 2xy), where A is parameter.
उत्तर
Differential equations
`dy/dx + (y^2 + y + 1)/(x^2 + x + 1)` = 0
or `dy/(y^2 + y + 1) + dx/(x^2 + x + 1) = 0`
and `dy/(y^2 + y + 1/4 + 3/4) + dx/(x^1 + x + 1/4 + 3/4) = 0`
or `dy/((y + 1/2)^2 + 3/4) + dx/((x + 1/2)^2 + 3/4)` = 0
On integrating,
`int dy/((y + 1/2)^2 + 3/4) + int dx/((x + 1/2)^2 + 3/4)` = 0
⇒ `2/sqrt3 tan^-1 ((y + 1/2)/(sqrt3/2)) + 2/sqrt3 tan^-1 ((x + 1/2)/(sqrt3/2))` = C
⇒ `2/sqrt3 tan^-1 ((2y + 1)/sqrt3) + 2/sqrt3 tan^-1 ((2x + 1)/sqrt3)` = C
⇒`2/sqrt3 tan^-1 [((2y + 1)/sqrt3 + (2x + 1)/sqrt3)/(1 - (2y + 1)/sqrt3 xx (2x + 1)/sqrt3)]` = C
⇒ `2/sqrt3 tan^-1 [(sqrt3 (2x + 2y + 2))/(3 - (2y + 1)(2x + 1))]` = C
⇒ `tan^-1 [(sqrt3(2x + 2y + 2))/(2 - 2x - 2y - 4xy)] = sqrt3/2`C
`= tan^-1 sqrt3 A`
Where C = `2/sqrt3 tan^-1 (sqrt3 A)`
`=> (2sqrt3 (x + y + 1))/(2 (1 - x - y - 2xy)) = sqrt3A`
∴ The required solution is
x + y + 1 = A(1 – x – y – 2xy)
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